a)$d(d(n)) = 2$
then we need to know that d(n) the amount of divisor is prime . $n = {p_1}^{k_1}*.......{p_n}^{k_n}$ total divisor = $(k_1+1)*....(k_n+1)$ hence it has only one factor since $(k_1+1)......(k_n+1)$ will make the divisor composite for $k_i$ natural
$n = (p_1)^{k_1}$ , $(k_1+1)$ is prime hence $k_1$ is even but still need to check
$n = (p_1)^{2k}$ for k natural
so n is a 2k-tic of a prime such as (quadratic,quartic,....>)
$2^{2k} = (4,16,64)$ 256 is not included since $2^8 = 9 \ divisor$
$3^{2k} = (9,81,729)$
$5^{2k} = (25,625)$
$7^{2k} = (49)$
$11^{2k}= (121)$
$13^{2k}= (169)$
$17^{2k} = ..$
$19^{2k} = ...$
$23^{2k}=.......$
$29^{2k}=.....$
$31^{2k} = (961)$
$(8+2+3+3) = 16$
pls correct me if there is something wrong