Answer is C.
Let the number of numbers divisible by 2 be denoted by $n(2);$ Numbers divisible by $3$ by $n(3);$ and the number of numbers divisible by $5$ by $n(5).$
Numbers divisible by 4 will be fully divisible by 2, so no need to consider those numbers separately.
$n(2) = 51$ $\left(\lfloor \frac{100}{2} \rfloor +1\right)$ [300 also included]
$n(3) = 33 \left(\lfloor \frac{100}{3} \rfloor\right)$
$n(5) = 21$ $\left(\lfloor \frac{100}{5} \rfloor +1\right)$ [300 also included]
$n(2 * 3) = 16 \left (\lfloor \frac{100}{2 \times 3} \rfloor \right)$
$n(3* 5) = 6 \left (\lfloor \frac{100}{3\times 5} \rfloor \right)$
$n(2*5) = 11 \left (\lfloor \frac{100}{2\times 5} \rfloor + 1\right)$ [300 also included]
$n(2 \cup 3 \cup 5) = n(2) + n(3) + n(5) - n(2*3) - n(3*5) - n(2*5) + n(2*3*5) = 51 + 33 + 21 – 16 – 6 – 11 + 3 = 75$ nos are divisible by 2, 3 or 5
Total nos = 100 + 1 = 101 [300 and 400 both included]
So, the number of numbers that are not divisible by $2,3,4$ and $5$ will be
$\qquad \text{total numbers } – n(2 \cup 3 \cup 5) = 101 – 75 = 26$