When face of the largest area is touching the table, height of centre of gravity above the table $h_1=\frac{6}{2}=3\text{cm}$

With its length vertical, height of centre of gravity would become $h_2=\frac{16}{2}=8\text{cm}$

Minimum work required W = (P.E.)₂ – (P.E.)₁

                                         = mg(h₂ – h₁)

$W=\frac{8\times 10\left(8-3\right)}{100}=4\text{J}$