The $2\times2$ and $1\times4$ tiles each have area $4$. Since the rectangle has area $21$, the number of $1\times1$ tiles must be congruent to $1\pmod4$. Among the choices, only $1$ and $5$ are possible by area.

It is impossible to use just one $1\times1$ tile. If the larger tiles covered the other $20$ cells, then two rows would have all $7$ cells covered by larger tiles. But each $2\times2$ tile covers $2$ cells in any row it meets, and each $1\times4$ tile covers $4$ cells in one row, so each row would have an even number of cells covered by larger tiles. A row cannot have $7$ such cells.

The following tiling shows that $5$ unit tiles are possible.

Thus, E is the correct answer.