48 should be the correct answer.

Let #@@@@ be the board where # denotes the leftmost strip and @'s denote remaining 4 strips.

We do not have to check for both left & right neighbour of each strip for boundary condition, but it would be enough to just check either left neighbour of each strip or right neighbour of each strip.

Here I am ensuring that none of the strip can have the same colour as its left neighbour.

Now since strip # has no left neighbour so it can have any of the 3 colours.

Remaining four strips that is @'s have exactly one left neighbour so, all of them have any of the 2 colours that is not possessed by their left neighbours, thus all of them will have 2 choices.

Which gives 3x2x2x2x2 = 48.