After reviewing the statements more carefully, I have deviced a method to determine the "real length" of a pencil (stick). If I can determine the values of the y (length) and x (time) factors, it will be simple to determine the pencil's length. In order to determine the y & x values, the following information is assumed/given: $g_1 = g_2 = g = 9.8 m/s^2$ ; $c_1 = c_2 = c$
I, and everything in the room shrunk by y = 1/2.
My thumb width tw, defined as 2cm long.
Solution: find out the shrink factor and thereby be able to find the length of the pencil.
Method: create a "pendulum clock" who's period is 1 second (local time). The length of a string for such pendulum is given by $L_2 = \frac{g{t_2}^2}{4{\pi}^2}$. For $T_2$ = 1 second, $L_2 = 24.85cm$. Using my thumb, I measure the string and create a "pendulum clock". Keep in mind that since I am shrunk by 1/2, $L_2$'s "real" length ($L_1$) is $\frac{L_2}{2}$, or 12.42cm. So,outside my shrunk room, this pendulum would have a period of $T_1 = 2 \pi$ $(\frac{L_1}{g})^{1/2} = \frac{1}{2^{1/2}}sec$. Using a fiber optic cable of length (d) of 3 x 10^10; an LED; and a light detector; the time delay should be $t_2$ = d/c = 1sec. However, the "real time" $t_1$ is = d/2c = 1/2 sec (since the real length is d/2), but since I am measuring time with my pendulum, then the pendulum period ($T_2$) required is $\frac{1/2}{\frac{1}{2^{1/2}}}$ = $\frac{1}{2^{1/2}}$. What must I do now to the string length so the pendulum's period is $\frac{1}{2^{1/2}}$ sec (instead of 1 sec)? I have to multiply its period by $\frac{1}{2^{1/2}}$, which means its length must be multiplied by 1/2, thus factor y = 1/2. So now, if you give me a stick that's 20cm long, I will determine it's 40cm long, but because I know the shrink factor is 1/2, I can say that the length of the stick is really 20cm.