Answer:
This solution is neutral.
Assumption: the solution is under room temperature, such that [tex]K_w = 10^{-14}[/tex].
Explanation:
For a solution in water,
[tex]\mathrm{[H_3O^{+}] \cdot [OH^{-}]} = \mathnormal{K_w}[/tex].
In other words, if [tex]K_w[/tex] is given, knowing the concentration (in [tex]\rm M[/tex], or equivalently [tex]\rm mol\cdot L^{-1}[/tex]) of either [tex]\rm H_3O^{+}[/tex] or [tex]\rm OH^{-}[/tex] will imply the concentration of the other ion.
Under room temperature,
[tex]K_w \approx 10^{-14}[/tex].
The question states that for this solution,
[tex]\mathrm{[H_3O^{+}]} = \rm 1.00\times 10^{-7}\; M[/tex]
As a result, the concentration of [tex]\rm OH^{-}[/tex] in this solution will be
[tex]\displaystyle \frac{K_w}{\mathrm{[H_3O^{+}]}} = \frac{10^{-14}}{1.00\times 10^{-7}} = 10^{-7}[/tex].
In this case, [tex]\rm [H_3O^{+}] = 10^{-7}\; M = [OH^{-}][/tex]. In other words, this solution is neutral.