To solve this problem, we can use the formula for osmotic pressure:

π = (n/V)RT

Where:
π is the osmotic pressure
n is the number of moles of solute
V is the volume of the solution
R is the ideal gas constant
T is the temperature in Kelvin

Let's assume that the number of moles of solute is constant in both cases. Therefore, we can write:

π1 = (n/V1)RT1
π2 = (n/V2)RT2

where π1 and π2 are the osmotic pressures at concentrations C and C/2 respectively, and temperatures 3270C and 4270C.

We are given that the osmotic pressure at concentration C/2 and temperature 4270C is 2 atm. Therefore, we can write:

2 = (n/V2)RT2

Now, we need to find the value of the osmotic pressure (P) at concentration C and temperature 3270C.

To do this, we can rearrange the equation for π1:

π1 = (n/V1)RT1

Now, let's substitute the given values:

π1 = (n/V1)RT1

π1 = (n/(2V2))RT1

We know that the temperature is 4270C, so we need to convert it to Kelvin:

T1 = 427 + 273 = 700K

Now, let's substitute the values into the equation:

π1 = (n/(2V2))RT1

π1 = (n/(2V2))(0.0821)(700)

We are given that the osmotic pressure at concentration C/2 and temperature 4270C is 2 atm. Therefore, we can write:

2 = (n/V2)(0.0821)(700)

Now, let's rearrange this equation to solve for n/V2:

(n/V2) = 2/(0.0821)(700)

(n/V2) = 0.036

Now, let's substitute this value back into the equation for π1:

π1 = (0.036)(n/(2V2))(0.0821)(700)

Simplifying this expression, we get:

π1 = 0.036π2

Therefore, the value of P is 0.036 times the osmotic pressure at concentration C/2 and temperature 4270C.

Since the osmotic pressure at concentration C/2 and temperature 4270C is given as 2 atm, we can calculate P:

P = 0.036 * 2
P = 0.072 atm

Therefore, the value of P is 0.072 atm or 72/100 atm, which can be simplified to 24/7 atm. Hence, the correct answer is option B (24/7).