Step 1: Use the key idea.
Freezing point depression depends only on molality for non-electrolytes, since $\Delta T_f = K_f\, m$. Both urea and glucose are non-electrolytes, so $K_f$ is the same.
Step 2: Molality of urea.
Urea has molar mass $60$. Moles $= 45/60 = 0.75$ in $1\,kg$ water, so $m_1 = 0.75\ mol\,kg^{-1}$.
Step 3: Molality of glucose.
Glucose has molar mass $180$. Moles $= 90/180 = 0.5$ in $2\,kg$ water, so $m_2 = 0.5/2 = 0.25\ mol\,kg^{-1}$.
Step 4: Set up the ratio.
Since $\Delta T_f \propto m$, \[ \frac{\Delta T_{f,2}}{\Delta T_{f,1}} = \frac{m_2}{m_1} = \frac{0.25}{0.75} = \frac{1}{3}. \]
Step 5: Find the new depression.
\[ \Delta T_{f,2} = \frac{1}{3}\times 1.395 = 0.465\,K. \]
Step 6: State the answer.
The freezing point of the glucose solution falls by $0.465\,K$.
\[ \boxed{0.465\,K} \]