We need to check that which of the given options has exactly three positive divisors.

The number having exactly three positive factors must be of the form $\(=Prime ^2\)$

( $\(\because\)$ The number of positive factors of a number $\(N\)$ of the form $\(N=a^p \times b^q \times c^r\)$, where $\(a, b \& c\)$ are distinct prime numbers, is $\((p+1)(q+1)(r+1))\)$

So, the options which are squares of a prime number is/are the correct answer/s.

(A) $\(121=11^2\)$, where 11 is a prime number, so 121 has exactly three positive divisors.
(B) $\(81=9^2\)$, where 9 is a composite number, so 81 doesn't have exactly 3 divisors.
(C) $\(1331=11^3\)$ which is not of square of prime form, so is incorrect.
(D) $\(49=7^2\)$ which is of the required form, so is correct.
(E) $\(729=9^3\)$ which is not of the required form, so is incorrect.

Hence options (A) \& (D) are correct.