It's Time To Take Sides

Relevant wiki: Perfect Squares, Cubes, and Powers

Clearly $N$ must be a multiple of both 2 and 3. Let $N={ 2 }^{ a }\cdot { 3 }^{ b }\cdot k$ .

Then $\frac { N }{ 2 } = { 2 }^{ a-1 }\cdot { 3 }^{ b }\cdot k$ . If this is a perfect square, $a-1$ and $b$ must be even, and $k$ must be a square.

Also, $\frac { N }{ 3 } = { 2 }^{ a }\cdot { 3 }^{ b-1 }\cdot k$ . Hence $a$ and $b-1$ must be multiples of 3, and $k$ must be a cube.

The smallest $a$ satisfying both of these conditions is 3, and the smallest $b$ is 4. For $N$ to be smallest, we choose $k=1$ , so $N={ 2 }^{ 3 }\cdot { 3 }^{ 4 } = \boxed { 648 }$ .