Answer
Cyclopentane will have the smaller amount of strain when forced to have all its carbons in a plane.
This is because in a plane, cyclopentane exists in pentagonal shape. For a regular pentagon, every bond angle is $108^{\circ}$. Now, If we look at the hybridization of every C atom in cyclopentane, every C atom is $sp^{3}$ hybridized. so, the C-C-C bond angle should be $109.5^{\circ}$ which is close to $108^{\circ}$. So, we can say it will face less strain when forced to have all its carbons in plane.
On the other hand, in a plane cyclohexane exists in hexagonal shape. For a regular hexagon, every bond angle is $120^{\circ}$. But C-C-C bond angle should be $109.5^{\circ}$. So, cyclohexane will face more strain than cyclopentane when forced to have all its carbons in plane.
Work Step by Step
In a plane, cyclopentane exists in pentagonal shape. For a regular pentagon, every bond angle is $108^{\circ}$. Now, If we look at the hybridization of every C atom in cyclopentane, every C atom is $sp^{3}$ hybridized. so, the C-C-C bond angle should be $109.5^{\circ}$ which is close to $108^{\circ}$. So, we can say it will face less strain when forced to have all its carbons in plane.
On the other hand, in a plane cyclohexane exists in hexagonal shape. For a regular hexagon, every bond angle is $120^{\circ}$. But C-C-C bond angle should be $109.5^{\circ}$. So, cyclohexane will face more strain than cyclopentane when forced to have all its carbons in plane.
