Approximately days have passed when 20% of cobalt (co) remains?

The half-life of Cobalt 60 is 5.27 years. This means that after this time, only 50 % of the initial amount of Cobalt is left. The relationship between the number of nuclei left at time t and the time, for a radioactive decay, is given by

where  is the number of nuclei left at time t,  is the number of nuclei at t=0, and  is the half-life. For cobalt, .

We can re-arrange the formula as

and then

the problem asks for the time t at which only 20% of cobalt is left, that means the time t at which

therefore, using this into the previous equation, we get

so, after a time of 12.24 years, only 20% of cobalt is left.

Ohm's law tells you that i = v/r where v is equivalent voltage and r is equivalent resistance of the circuit
since all circuits have the same voltage, your question is really asking which circuit has the lowest total resistance, since the lower the resistance, the higher the current
circuit a has a resistance of 5 ohms; circuit c is in series so has a resistance of 10 ohms; circuit b is in parallel and therefore has a resistance of 2.5 ohms (review your notes on adding resistances in parallel and series); therefore b has the highest current