$\large\frac{dV_A}{\sigma t}$$=-K_1 V_A$

=> $\large\frac{dV_A}{V_A}=-k_1 dt$

$\log V_A=-K_1t+c$

at $t=0\quad V_A=V_0 A \quad \log V_{OA}=c$

$\log \bigg( \large\frac{VA}{V_{OA}}\bigg)$$=-K_1t$

$=> V_A=V_OA e^{-K_1t}$

Similarly for reservoir B $ VB =V_oB e^{-K_2t}$

$VOA=2 VOB$

$V_B=\large\frac{VOA}{2} e^{-K_2 t}$

at $t= 1 hours$

$V_A= \large\frac{3}{2} $$V_B$

$VOA e^{-K_1} =\large\frac{3}{2} \large\frac{V_o Ae^{-K_2}}{2}$

$e^{k_2-K_1}=\large\frac{3}{4}$

$K_1 -K_2= \log \bigg(\large\frac{4}{3}\bigg)$

at $t=t$

$V_A=V_B$

$VOAe^{-K1t}=\large\frac{VOA}{2} e^{-k_2t}$

$2= e^{(K_1-K_2)t}$

$e^{(K_2-K_1)t}= \large\frac{1}{2}$

$(K_2-K_1)t=-\log 2$

$t= \large\frac{\log 2}{K_1-K_2}$

$t= \large\frac{\log 2}{\log 4/3}$

$\quad=\large\frac{\log 2}{\log 4- \log 3}$

$a=2,b=4,c=3$

$a+b-c=3$

Hence b is the correct answer.