In base $'B'$ number system the number,
$(201)_{B} = (1 \times B^{0}) + (0 \times B^{1}) + (2\times B^{2})$
$\implies(201)_{B} = 2B^{2} + 1\qquad \rightarrow(1)$
and, $(401)_{B} = (1\times B^{0}) + (0 \times B^{1}) + (4 \times B^{2})$
$\implies (401)_{B} = 4B^{2} + 1 \qquad \rightarrow(2)$
Subtract equation $(2$ from equation $(1),$ we get
$4B^{2} + 1 -(2B^{2} + 1) = 128$
$\implies 2B^{2} = 128$
$\implies B^{2} = 64$
$\implies B = 8,\, B \neq -8,$ because negative base is not possible.
Thus base of the system $B = 8.$
Now, we need to find $(2424)_{8} = (\,)_{10}$
$\implies (2424)_{8} = (4 \times 8^{0}) + (2 \times 8^{1}) + (4 \times 8^{2}) + (2 \times 8^{3})$
$\implies (2424)_{8} = 4 + 16 + 256 + 1024 = (1300)_{10}.$
So, the correct answer is $(C).$