From the above figure, we can see that due to mass placed at edge of table, it tends to rotate about a line joining two legs. Hence, we take pivot in balancing of weight of mass $m$ and weight of mass $M$ of table at point $B$. In critical condition,

Moment of weight of mass $m$ about $B=$ Moment of weight of table about $B$

$\Rightarrow mg·\left(AB\right)=Mg·\left(BC\right)$

$\Rightarrow m=\frac{M·\left(BC\right)}{AB}$........(i)

From geometry of table,

$BC=R\cos 45°=\frac{R}{\sqrt{2}}$

So, $AB=R-\frac{R}{\sqrt{2}}$

$=R\left(1-\frac{1}{\sqrt{2}}\right)$

Hence, from Eq. (i), we have

$m=\frac{M\times R/\sqrt{2}}{R\left(1-\frac{1}{\sqrt{2}}\right)}$

Here, $M=20 \mathrm{kg}$

$\therefore m=\frac{20}{\sqrt{2}-1}\approx \frac{20}{0.4}\approx 50 \mathrm{kg}$

Nearest value is $m=47 \mathrm{kg}$.