The initial velocity for the reaction will be #"350 nmol s"^(-1)#.

Start from the **Michaelis-Menten equation**

#V_0 = (V_"max" * [S])/(V_"max" + [S])#

When an **uncompetitive inhibitor** is introduced, the equation takes this form

#V_0 = (V_"max" * [S])/(K_m + [S] * underbrace((1 + ([I])/K_I))_(alpha)#

But #(1 + ([I])/K_I)# is actually equal to #alpha#, the *degree of inhibition*, which implies that the equation becomes

#V_0 = (V_"max" * [S])/(K_m + [S] * alpha)#

Now plug your values and solve for #V_0#

#V_0 = ("950 nmol s"^(-1) * 350cancel(mu"mol L"^(-1)))/((175+ 350 * 2.20)cancel(mu"mol L"^(-1))#

#V_0 = 350/(945) * "950 nmol s"^(-1) = "351.85 nmol s"^(-1)#

Rounded to two sig figs, the number of sig figs given for #V_"max"# and #[S]#, the answer will be

#V_0 = color(green)("350 nmol s"^(-1))#