Finding $E_b/N_0$ when coding rate is $R_c$
The energy per bit, $E_b$, is independent of the coding rate. Note that $E_b$ measures the energy per transmitted information bit, not per transmitted symbol.

Let's say you're willing to spend one joule per information bit, so $E_b=1$. You use uncoded BPSK, so that each transmitted symbol carries one bit of information and so it also has energy one. Let us say that $E_s=1$, where the subscript $s$ denotes "symbol". Let us further assume that you're operating at an SNR per bit of 10 dB; that is, $E_b/N_0=1/0.1=10$.

Now you decide to try a repetition code, with three repetitions per bit. With this code, a bit 1 is transmitted as three consecutive positive symbols, and a bit 0 as three negative symbols. If you're unwilling to increase the energy per bit, then each symbol must have energy equal to 1/3 of a joule: $E_s=1/3$. In this way, you still have $E_b=3E_s=1$.

Now, in the channel, the SNR of each symbol will be reduced: since each symbol has energy 1/3, then each symbol has SNR $E_s/N_0 = (1/3)/0.1 = 5.2 \text{ dB}$. However, the SNR per information bit hasn't changed; it remains  $E_b/N_0=1$.