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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 $(C).$
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