$W = 4\, eV$
$KE_{max} = 6\, eV = KE_{initial}$
On applying accelerating potential of $7.6\,V$
$\Delta KE = charge × potential = 7.6\, eV$
$KE_f -KE_g = 7.6\, eV$
$KE_{final} = 7.6\, eV + 6\, eV = 13.6\, eV$
which is $KE$ of electron in $(n = 1)$ in hydrogen atom
$V = 2.188 \times 10^6\, m/s$
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$\begin{array}{*{20}{c}}
{C{H_3} - C = CH - C{H_2}C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{CH{{(C{H_3})}_2}\,\,\,\,\,\,\,\,\,\,}
\end{array}\xrightarrow[{(ii)\,{H_2}{O_2},O{H^ - }}]{{(i)\,{B_2}{H_6}}}[A]$$\xrightarrow[\Delta ]{{dil.\,{H_2}S{O_4}}}[B]$