Determine the equilibrium constant for this reaction
You made a slight error in calculations for B and C. While those concentrations are correct, that is 1 M at 15 mL. When doing RICE tables and equilibrium calculations, we can use concentrations, however I prefer to convert concentration into moles.

We are trying to find $K_c$, and we know that equal moles of the reactant react to give us 0.0045 mol of BC.

$$\ce{B + C &lt;=&gt; BC}$$

\begin{array} {|c|c|c|c|}
\hline
\text{Initial conc.} &amp; \text{0.015 mol} &amp; \text{0.015 mol} &amp; \text{0 mol}\\
\hline
\text{Change conc.} &amp; -x &amp; -x &amp; +x\\
\hline
\text{End conc.} &amp; \text{0.015 mol} - x &amp; \text{0.015 mol} - x &amp;  0 + x \\
\hline
\end{array}

We know that our change in equilibrium is 0.00450 mol, as that is the end amount of product that we have in our equilibrium calculations.

Accounting for the concentrations, we divide our number of moles by the total volume to get mol/L. 0.5 mol/L, 0.5 mol/L, .15 mol/L. 

Assuming that our reaction is at equilibrium:

$$K_c = \frac{[\text{0} + \text{.15 mol/L}]}{[\text{0.5 mol/L} - \text{.15 mol/L}][\text{0.5 mol/L} - \text{.15 mol/L}]} $$

$$K_c = \frac{[\text{.15 mol/L}]}{\text{[0.35 mol/L]}^2} $$

$$ K_c = 1.22 $$

Thanks @Nicolau for pointing out the error in the question and reviewing my work.