#### Answer

(a) Less than
(b) $1.283Kg$

#### Work Step by Step

(a) We know that
$Q=mc\Delta T$
This can be rearranged as:
$m=\frac{q}{c\Delta T}$
As $Q$ and $\Delta T$ are same for aluminum and copper
$\implies m\propto \frac{1}{c}$
We know that $c_{Al}=0.900J/gC^{\circ}$
and $c_{Cu}=0.385J/gC^{\circ}$
which means $c_{Al}\gt c_{Cu}$ therefore $m_{Al}\lt m_{Cu}$
We conclude that the mass of the aluminum block is less than that of copper.
(b) We know that
$m_{Al}c_{Al}=m_{Cu}c_{Cu}$
This can be rearranged as:
$m_{Al}=(\frac{c_{Cu}}{c_{Al}})m_{Cu}$
We plug in the known values to obtain:
$m_{Al}=(\frac{0.385}{0.900})(3)=1.283Kg$