Answer
$$26.26\;m^2$$
Work Step by Step
Let, $A$ be the area of the top surface of the ice slab and $h$ be the thickness. So volume of the ice slab is $V=Ah$. We assume that the top surface of the ice slab is at the surface of water and it holds up an automobile of mass $M$
Now, the total downward force of gravity on the ice and automobile is:
$W=(M + \rho_i Ah)g$
The buoyant force of the water is acted only on the ice slab, which is given by
$F_b=\rho_wAhg$
At equilibrium condition:
$W=F_b$
or, $(M + \rho_i Ah)g=\rho_wAhg$
or, $Ah(\rho_w-\rho_i)=M$
or, $A=\frac{M}{h(\rho_w-\rho_i)}$
Substituting the given values
$A=\frac{938}{0.441\times(998-917)}\;m^2$
or, $\boxed{A=26.26\;m^2}$
The minimum area of the top surface of the ice slab is $26.26\;m^2$
