1. Find the molar mass of $\mathrm{CCl_4}$

$M(\mathrm{CCl_4}) = 12 + 4(35.5) = 12 + 142 = 154\ \mathrm{g\,mol^{-1}}$

2. Calculate number of moles in $284\ \mathrm{g}$

$n = \frac{284}{154} = \frac{142}{77} \approx 1.844\ \mathrm{mol}$

3. Use enthalpy of vapourisation

Given:

$\Delta H_{vap} = 30.5\ \mathrm{kJ\,mol^{-1}}$

Heat required:

$q = n\,\Delta H_{vap}$

$q = \frac{284}{154}\times 30.5$

Since

$30.5 = \frac{61}{2}, \qquad \frac{284}{154}=\frac{142}{77}$

So,

$q = \frac{142}{77}\times \frac{61}{2} = \frac{71\times 61}{77}$

$q = \frac{4331}{77} \approx 56.25\ \mathrm{kJ}$

4. Final integer answer

$\boxed{56}$

Thus, the heat required for vapourisation of $284\ \mathrm{g}$ of $\mathrm{CCl_4}$ is $56\ \mathrm{kJ}$ (approximately).