Let the initial velocity of the bullet be $u$.

After penetrating $4\text{ cm}$ into the block, its velocity becomes $v = \frac{u}{3}.$

Since the resisting force is constant, acceleration is constant.

So we can use the kinematic equation: $v^2 = u^2 + 2as$

For the first $4\text{ cm}$ of motion: $\left(\frac{u}{3}\right)^2 = u^2 + 2a(4)$

$\frac{u^2}{9} = u^2 + 8a$

$8a = \frac{u^2}{9} - u^2 = -\frac{8u^2}{9}$

$a = -\frac{u^2}{9}$ (Here distance is in cm, so acceleration is in consistent units.)

Now find the total distance travelled before coming to rest.

When the bullet stops completely, final velocity $=0$.

Again using $v^2 = u^2 + 2as$

$0 = u^2 + 2\left(-\frac{u^2}{9}\right)s$

$0 = u^2 - \frac{2u^2}{9}s$

$\frac{2}{9}s = 1$

$s = \frac{9}{2} = 4.5\text{ cm}$

Given total distance is $(4+x)$ cm:

$4 + x = 4.5$

$x = 0.5$

Match with options:

The correct option is: $\boxed{\text{C: }0.5}$