Given,

$X=$ Number of white ball drawn,

And bag contains $4$ white and $6$ black balls, total $3$ balls are to be drawn,

So, $P\left(X=0\right)=\frac{{}^{6}C_3}{{}^{10}C_3}=\frac{1}{6}$

$P\left(X=1\right)=\frac{{}^{6}C_2\times {}^{4}C_1}{{}^{10}C_3}=\frac{1}{2}$

$P\left(X=2\right)=\frac{{}^{6}C_1\times {}^{4}C_2}{{}^{10}C_3}=\frac{3}{10}$

and $P\left(X=3\right)=\frac{{}^{6}C_0\times {}^{4}C_3}{{}^{10}C_3}=\frac{1}{30}$$$

$X$	$0$	$1$	$2$	$3$
$P\left(X\right)$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{3}{10}$	$\frac{1}{30}$

So, variance $={\sigma }^2=\sum P_i{X}_i^2-{\left(\sum P_i{X}_i\right)}^2$

$\Rightarrow {\sigma }^2=\frac{1}{2}+\frac{12}{10}+\frac{3}{10}-{\left(\frac{1}{2}+\frac{6}{10}+\frac{1}{10}\right)}^2$

$\Rightarrow {\sigma }^2=\frac{56}{100}$

$\Rightarrow 100{\sigma }^2=56$