Answer: 109

Small Hint:

A number has exactly four divisors in all — hence three proper ones — exactly when it is $pq$ for distinct primes $p$ and $q,$ or a prime cube $p^3$

Big Hint:

Count pairs of distinct primes less than $50,$ then add the primes $p$ with $p^2 \lt 50$

Solution:

An integer with exactly three proper divisors has exactly four divisors in total, so it is either $n = pq$ with $p$ and $q$ distinct primes (proper divisors $1,$ $p,$ $q$) or $n = p^3$ with $p$ prime (proper divisors $1,$ $p,$ $p^2$).

In the first case we need $p$ and $q$ both less than $50.$ There are $15$ primes below $50,$ giving $\binom{15}{2} = 105$ such numbers. In the second case we need $p^2 \lt 50,$ which holds for $p = 2,$ $3,$ $5,$ $7,$ giving $4$ more.

The total is $105 + 4 = 109.$