Write n as 100x + 10a + b where a and b are integers < 10 and x may be > 10. n^2 = 10000x^2 + 100a^2 + b^2 + 2000xa + 200xb + 20ab, and we can ignore terms with multiple of 100 since they can't affect the final two digits. That means the last two digits are controlled by 20ab + b^2, whose tens digit is given as 7. (And only b^2 determines the final digit.)

The ab term's contribution is necessarily even, and 7 is odd, so the b term must provide an odd contribution. The only squares of single digits with odd tens digits are 16 = 4^2 and 36 = 6^2, so b must be 4 or 6. Regardless of which value of b, b^2 ends in 6, so the last digit is 6.