Solution:

The answer is (A). Squaring the equation in the problem we obtain the equivalent condition $a + 2 \sqrt{a b} + b = a + c \sqrt{b}$, from which, subtracting $a$ and dividing by $\sqrt{b}$, we arrive at
$$2 \sqrt{a} + \sqrt{b} = c.$$
For $c$ to be an integer, both $a$ and $b$ must be perfect squares. This can be observed by squaring the equality $2 \sqrt{a} = c - \sqrt{b}$, from which we obtain $4a = c^2 + b - 2c \sqrt{b}$; therefore $\sqrt{b}$ is a rational number, so $b$ is a perfect square, and in particular $\sqrt{b}$ is an integer. Going back to the starting equation $2 \sqrt{a} + \sqrt{b} = c$, one immediately deduces that $a$ must also be a perfect square.

Given the condition $a, b \leq 2021$, we therefore need to count how many values the expression $2x + y$ takes with $1 \leq x, y \leq 44$.

It is easy to see that all and only the values between $3$ and $3 \cdot 44$ inclusive are attained: with $x = 1$ the interval $[3, 46]$ is filled, with $x = 23$ the interval $[47, 88]$ is taken, and finally with $x = 44$ the numbers in the interval $[89, 132]$ are also reached.