Lumped Capacitance

Consider a 2 m long metal cube has a convection heat transfer coefficient of 750 W/m^2∙K, and a thermal conductivity of 500 W/m∙K. Is the lumped capacitance model valid for this part?

Hint
$$$Bi=\frac{hV}{kA_s}$$$
where $$Bi$$ is the Biot Number, $$h$$ is the convection heat transfer coefficient, $$V$$ is the volume, $$k$$ is the thermal conductivity, and $$A_s$$ is the surface area.
Hint 2
The lumped capacitance model is valid if $$Bi< 0.1$$ .
The lumped capacitance model is valid if
$$$Biot\:Number, Bi=\frac{hV}{kA_s}< 0.1$$$
where $$h$$ is the convection heat transfer coefficient, $$V$$ is the volume, $$k$$ is the thermal conductivity, and $$A_s$$ is the surface area.
$$$Bi=\frac{750W(2m)^3(m\cdot K)}{(m^2\cdot K)(500W)(6\cdot (2m)^2)}=\frac{6,000}{12,000}=0.5$$$
Since $$Bi>0.1$$ , the Lumped Capacitance Model is not valid .
No