The correct answer is **B) $20 \\frac{260}{587}$ days**.\n\nHere's a detailed step-by-step solution:\n\n**Step 1: Determine Individual Efficiencies and Total Work**\nLet the total work be $L$ units.\n\n* **Efficiency of B ($E_B$):** B completes the work in 60 days. So, B's 1-day work is $\\frac{L}{60}$ units/day.\n $E_B = \\frac{L}{60}$\n\n* **Efficiency of A ($E_A$):** A is 20% more efficient than B.\n $E_A = E_B \\times (1 + 0.20) = 1.2 \\times E_B = 1.2 \\times \\frac{L}{60} = \\frac{1.2L}{60} = \\frac{L}{50}$ units/day.\n\n* **Efficiency of C ($E_C$):** C is 25% less efficient than A.\n $E_C = E_A \\times (1 - 0.25) = 0.75 \\times E_A = 0.75 \\times \\frac{L}{50} = \\frac{3}{4} \\times \\frac{L}{50} = \\frac{3L}{200}$ units/day.\n\nTo simplify calculations, let's assume a convenient value for the Total Work, $L$. We choose $L$ as the Least Common Multiple (LCM) of the denominators (60, 50, 200), which is 600.\n\nSo, let **Total Work = 600 units.**\n\n* $E_B = \\frac{600}{60} = 10$ units/day.\n* $E_A = \\frac{600}{50} = 12$ units/day. (Check: $12 = 1.2 \\times 10$, correct).\n* $E_C = \\frac{3 \\times 600}{200} = \\frac{1800}{200} = 9$ units/day. (Check: $9 = 0.75 \\times 12$, correct).\n\n**Step 2: Calculate Work Done in Phase 1 (A and B for 5 days)**\n* Combined efficiency of A and B ($E_{A+B}$) = $E_A + E_B = 12 + 10 = 22$ units/day.\n* Work done in 5 days = $22 \\text{ units/day} \\times 5 \\text{ days} = 110$ units.\n* Remaining Work = Total Work - Work done in Phase 1 = $600 - 110 = 490$ units.\n\n**Step 3: Calculate Work Done in Phase 2 (B and C at 80% efficiency for 6 days)**\n* C's reduced efficiency ($E_{C\_red}$) = $E_C \\times 80\\% = 9 \\times 0.8 = 7.2$ units/day.\n* Combined efficiency of B and C (reduced) ($E_{B+C\_red}$) = $E_B + E_{C\_red} = 10 + 7.2 = 17.2$ units/day.\n* Work done in 6 days = $17.2 \\text{ units/day} \\times 6 \\text{ days} = 103.2$ units.\n* Remaining Work = Work remaining after Phase 1 - Work done in Phase 2 = $490 - 103.2 = 386.8$ units.\n\n**Step 4: Calculate Efficiency of D (Phase 3)**\n* In Phase 3, B, C (at normal efficiency), and D complete the remaining work in 8 days.\n* Let D's efficiency be $E_D$ units/day.\n* Combined efficiency of B, C (normal), and D ($E_{B+C+D}$) = $E_B + E_C + E_D = 10 + 9 + E_D = (19 + E_D)$ units/day.\n* Work done in 8 days = $(19 + E_D) \\times 8$ units.\n* This work must be equal to the remaining work from Phase 2:\n $(19 + E_D) \\times 8 = 386.8$\n $152 + 8E_D = 386.8$\n $8E_D = 386.8 - 152$\n $8E_D = 234.8$\n $E_D = \\frac{234.8}{8} = \\frac{117.4}{4} = \\frac{58.7}{2} = 29.35$ units/day.\n\n**Step 5: Calculate Days D Alone Takes to Complete the Entire Work**\n* Days = Total Work / D's efficiency\n* Days D takes = $\\frac{600 \\text{ units}}{29.35 \\text{ units/day}}$\n* To get rid of the decimal, multiply numerator and denominator by 100:\n Days D takes = $\\frac{60000}{2935}$\n* Simplify the fraction by dividing by 5:\n Days D takes = $\\frac{12000}{587}$\n\nNow, convert this improper fraction to a mixed number:\n* Divide 12000 by 587:\n $12000 \\div 587 = 20$ with a remainder.\n $587 \\times 20 = 11740$\n Remainder $= 12000 - 11740 = 260$\n\nSo, D alone would take $20 \\frac{260}{587}$ days to complete the entire work.\n\nFinal Answer is **B) $20 \\frac{260}{587}$ days**