How do I set the speed of a 4060B chip?
According to the datasheet the formula for the R/C oscillator is:

\$ {\dfrac{1}{2.3 \cdot R1\cdot Cx}} \$

So for R1 = 10k\$ \Omega \$ and Cx = 10\$\mu F \$ 

\$ {\dfrac {1}{2.3 \cdot 10k \Omega\cdot 10 \mu F}} = 4.34Hz \$

You can use any of the Q4 to Q14 pins for output, they have different division ratios of the oscillator speed.

Where Osc = the oscillator frequency the frequency of each Q pin is Q4 = Osc / 16, Q5 = Osc / 32, Q6 = Osc / 64 and so on up to Q14 = Osc / 16384.   

So with the above example Q4 will toggle every \$ {\dfrac{1}{4.34Hz}} \cdot 16 = 3.68\$  seconds

For five minutes you simply need to choose a compatible frequency and divider ratio. 5 * 60 = 300 seconds. If we choose the divider as Q6 then 300/64 = 4.68 seconds needed for the oscillator. 

A quick shuffle of some figures gives one possible way as R1 = 204k\$\Omega\$ and Cx as 10\$\mu\$F. This would give:  

\$ 64 \cdot \left( \dfrac{1} {{\dfrac{1}{2.3 \cdot 204k\Omega \cdot 10\mu F} }}\right) = 64 \cdot 2.3 \cdot  204k\Omega \cdot 10\mu F = 300.288\$ seconds. 

Pretty close. I would probably use a smaller more precise capacitor and a larger resistor for more accurate timing. For best accuracy use the crystal option.