For an isothermal free expansion of an ideal gas correct option is :

$\mathrm{\Delta }U=q=w=\mathrm{\Delta }H=0$

For an adiabatic process, $q=0$, and for free expansion of a gas, $w=0$

$\therefore \mathrm{\Delta }U=q+w=0+0=0$

For an ideal gas undergoing an isothermal process, $∆U=0$. So, the given process is isothermal and hence $∆T=0$.