The volume of nickel plated is surface area times the thickness.
That's $2m^2 \times 0.03 \ cm $
We'll use cubic cm as the units so it's $2m \times 1m \times 0.03 \ cm $ = $200 cm \times 100 cm \times 0.03 \ cm= 600 \ cm^3 $

This helps to find the mass as the density is given as $8.9 g/cm^3$.
The mass = density $\times$ volume = $ 8.9 \ g/cm^3 \times 600 \ cm^3 = 5.34 \ kg$

Convert this to moles of Nickel. $\displaystyle \frac{5.34 \times 10^3 \ g}{58.71 \ g/mol} = 90.96 \ mol $
We'll come back to this. 

Now for the current, $0.3 A/cm^2$ applied to a 5 cm square plate, meaning that the total current on the plate at a given time is $0.3 A \times 25 \ cm^2 = 7.5 \ A$ or $7.5 \ C/s$. However, only a fraction of that actually reduces the nickel (75%) so $5.625 \ A$

The voltage is then $ V = IR = 5.625 \times 0.4 $ 

The time calculation involves the actual nickel.
$I = \displaystyle \frac{q}{t}$  
Where q = total charge

We have 90.96 mols of Nickel reduced.

$Ni^{2+} + 2e^-$ -> $Ni$

That's 2 moles of electrons per mole of Nickel. So we need $181.92$ mol of electrons.
The charge per mol of electrons is Faraday's constant of $96485.3 \ C/mol$
The total charge is just $181.92 mol \times 96485.3 \ C/mol$

Then we rearrange the formula of current for time $\displaystyle t = \frac{q}{I} = \frac{181.92 mol \times 96485.3 \ C/mol}{5.625 C/sec} $
$time = 3120463 \ seconds$ or $866 \ hours$

That's sort of the basic idea I'd imagine. I may have made a mistake somewhere in between. The time though while seems long is in line with this:
https://sciencing.com/calculate-electroplating-7597391.html

There, they plated just one mole of Cu and with a higher current. That took 2 hours. We have a lot more of Ni and smaller current.

Then $Energy = power \times time = VIt$ and you can calculate the cost from there.