From the relation
$h=ut+\frac{1}{2}gt^2$
$h=\frac{1}{2}gt^2 \Rightarrow g=\frac{2h}{t^2}$ ($\because$ body initially at rest)
Taking natural logarithm on both sides, we get
In $g = ln\, h - 2 \,ln\, t$
Differentiating, $\frac{\Delta g}{g}=\frac{\Delta h}{h}-2\frac{\Delta t}{t}$
For maximum permissible error.
or $\big(\frac{\Delta g}{g}\times100 \big)_{max}=\big(\frac{\Delta h}{h}\times100 \big)+2\times \big(\frac{\Delta t}{t}\times100 \big)$
According to problem
$\frac{\Delta h}{h}\times100 =e_1$ and $\frac{\Delta t}{t}\times100$
$ =e_2$
Therefore,$\big(\frac{\Delta g}{g}\times100 \big)_{max}$
$=e_1+2e_2$