\n\nThe correct answer is **B**. \n\nTo find the remainder when $7^{7^{77}}$ is divided by $10$, we need to analyze the powers of $7$ modulo $10$.\n\n**Step 1: Determine the cyclicity of the base modulo the divisor.**\nThe powers of $7$ modulo $10$ follow a cycle:\n$7^1 \\equiv 7 \\pmod{10}$\n$7^2 = 49 \\equiv 9 \\pmod{10}$\n$7^3 = 7^2 \\times 7 \\equiv 9 \\times 7 = 63 \\equiv 3 \\pmod{10}$\n$7^4 = 7^3 \\times 7 \\equiv 3 \\times 7 = 21 \\equiv 1 \\pmod{10}$\n$7^5 \\equiv 7 \\pmod{10}$\n\nThe cycle length for the powers of $7$ modulo $10$ is $4$. This means that for any integer $k$, $7^k \\equiv 7^{k \\pmod 4}$ when the exponent $k$ is greater than or equal to the cycle length. If $k \\pmod 4 = 0$, we use $4$ as the effective exponent (since $7^4 \\equiv 1 \pmod{10}$). More precisely, we are interested in $7^E \\pmod{10}$ where $E = 7^{77}$. We need to find $E \pmod 4$.\n\n**Step 2: Calculate the exponent modulo the cycle length.**\nNow we need to find the remainder of $7^{77}$ when divided by $4$. We can do this by examining powers of $7$ modulo $4$.\n$7 \\equiv 3 \\pmod{4}$\nAlternatively, it's often simpler to use negative remainders when applicable: $7 \\equiv -1 \\pmod{4}$.\n\nSo, $7^{77} \\equiv (-1)^{77} \\pmod{4}$.\nSince $77$ is an odd number, $(-1)^{77} = -1$.\nTherefore, $7^{77} \\equiv -1 \\pmod{4}$, which is equivalent to $7^{77} \\equiv 3 \\pmod{4}$.\n\n**Step 3: Substitute the result back into the original expression.**\nWe found that $7^{77} \\equiv 3 \\pmod{4}$. This means the effective exponent for the original problem is $3$. So, $7^{7^{77}}$ is equivalent to $7^3$ in terms of its remainder when divided by $10$.\n\nThus, $7^{7^{77}} \\equiv 7^3 \\pmod{10}$.\n\n**Step 4: Calculate the final remainder.**\n$7^3 = 7 \\times 7 \\times 7 = 49 \\times 7 = 343$.\n$343 \\equiv 3 \\pmod{10}$.\n\nHence, the remainder when $7^{7^{77}}$ is divided by $10$ is $3$.