The sum of the perfect square between 120 and 300 is:

Sum of the squares of n consecutive numbers =

The sum of the perfect square between 120 and 300 = $$11^2+12^2+13^2+14^2+15^2+16^2+17^2$$

$$=\frac{17\left(17+1\right)\left(2\left(17+1\right)\right)}{6}-\frac{10\left(10+1\right)\left(2\left(10\right)+1\right)}{6}$$

$$=\frac{17\left(18\right)\left(35\right)}{6}-\frac{10\left(11\right)\left(21\right)}{6}$$

$$=51\times35-11\times35$$

$$=35\left(51-11\right)$$

$$=35\left(40\right)$$

$$=1400$$

Hence, the correct answer is Option A