It's just a simple cube!

The total number of cubes will be $n^{3}$ ,where n is the number of sides. If one face is coloured then $n^{2}$ cubes are coloured. If two adjacent faces are coloured then 2 $n^{2}$ - n cubes are coloured. Hence the expression for total uncoloured faces can be written as $n^{3}$ - $(kn^{2}$ -cn),where k represents the number of sides coloured and c represents the number of adjacent pairs.

Hence $n^{3}$ -k $n^{2}$ + cn =45

But 45 = $3 \times$ $3 \times 5$ .

Hence the required cubic eqn. can be of the forms

(i) $n^{2}$ $\times$ ( n + 2 ) = $n^{3}$ + 2 $n^{2}$ =45 . Here k = -2 which cannot be allowed.

(ii) $( n - 1 )^{2}$ $\times$ ( n + 1 ) = $n^{3}$ - $n^{2}$ -n +1 =45 .Here due to the presence of constant term 1, the eqn. is not of the required form.

(iii) $( n - 2 )^{2}$ $\times$ ( n ) = $n^{3}$ - 4 $n^{2}$ + 4n =45

Similarly $( n - 1 )^{2}$ $\times$ ( n + 1 ), $( n - 1 )^{2}$ $\times$ ( n + 1 ) etc. can be written but it will not be in the form of required expression as we have seen in case (ii).

Thus only case (iii) will satisfy the problem and hence k=4, c=4 and n=5. Thus the number of faces coloured is 4.