Explanation

We know that $\ \ \ \ \ \ \ \ \ \ \ \ \ Q = \frac{KA\left( \theta_{1} - \theta_{2} \right)t}{d}$ $\Rightarrow \ \ \ \ \ \ K = \frac{Qd}{A\left( \theta_{1} - \theta_{2} \right)t}$ So, the unit of $K = \frac{cal\ \times metre}{{metre}^{2}\ \times \ K\ \times \ sec}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{cal}{m}.K\ .\sec\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$ And density $d = \frac{m}{V},$ where $m =$ mass, $V =$ volume. So, the unit of $d = \frac{kg}{m^{3}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$ Also, we know that $Q = mc\mathrm{\Delta}\theta$ $\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ c = \frac{Q}{m\ .\ \mathrm{\Delta}\theta}$ So, the unit of $c = \frac{cal}{kg\ .K}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(iii)$ Hence, the unit of $X = \frac{K}{dc}$ On putting the values of unit from Eqs. (i), (ii) and (iii), the unit of $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ X = \frac{cal/mK - sec}{{kg/m}^{3} \times cal/kg\ .K}$ $\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = m^{2}/s$ So, the unit of $X$ in CGS system is ${cm}^{2}s^{- 1}.$